We are being asked to:
(a) determine the limiting reactant when 13.50 g of CH3CH3 reacts with 2.00 g of O2
(b) determine the amount of water formed when 13.50 g of CH3CH3 reacts with 2.00 g of O2.
The molar masses of the involved compounds solved below before proceeding with further calculations.
CH3CH3 2C × 12.01 g/mol = 24.02 g/mol
6H × 1.008 g/mol = 6.048 g/mol
SUM = 30.068 g/mol
O2 2O × 16.00 g/mol = 32.00 g/mol
SUM = 32.00 g/mol
CO2 1C × 12.01 g/mol = 12.01 g/mol
2O × 16.00 g/mol = 32.00 g/mol
SUM = 44.01 g/mol
H2O 2H × 1.008 g/mol = 2.016 g/mol
1O × 16.00 g/mol = 16.00 g/mol
SUM = 18.016 g/mol
For the following equation:
2CH3CH3 + 7O2 → 4CO2 + 6H2O
Answer these questions:
a) If 13.50g of CH3CH3 reacts with 2.00g of oxygen, which is the limiting reagent?
b) If 13.50g of CH3CH3 reacts with 2.00g of oxygen, calculate the number of g of water formed?
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